It's X.

The square root of x^2 is +/- sqrt(x^2) where sqrt gives you the principal square root. The square root of x^2 is +/- |x| if x is real. The square root of x^2 can also be written (-1)^k x, where k is some unknown integer.

X times X = X squared.

X^.5= +/- x presuming that x is a real number.

Please note than when X is real, X^2 is never negative. In general, for any nonnegative real number A there are two numbers, one nonnegative and one negative, that upon squaring produce A. For instance, if A=9 then 3 and –3 produce 9 upon squaring; in case of A=0 both numbers are equal to 0. Therefore, there are two numbers, X and –X, that upon squaring produce X^2. Please notice that you don't know which one is negative; it is not necessarily –X since what if X=–3? Usually we show the answer by putting both + and – in front of X, either separated by a slash or placing + on top of –. This should be interpreted as two actual possibilities. However, if you use the conventional symbol for the square root, that is / (the radical; I will use the common abbreviation sqrt since this site does not allow non-keyboard symbols), then there is a convention in math that it always mean the non-negative one of the two possible values. Thus, sqrt(9)=3, even though –3 squared also produces 9. The reason for the convention is to avoid ambiguity in algebraic notation by not allowing any symbols to generate more than one value. In elementary algebra we do not deal with functions that produce more than one value of Y per one value of X; multi-valued functions like those in complex analysis or in vector analysis are too complex conceptually. So, for sqrt(X^2), according to the convention, of X and –X we should choose the one that is non-negative. However, we cannot say which one is non-negative, since the letters denote variables, not particular numbers! The solution is to use the symbol for the absolute value | |, which always converts a real number or variable to something nonnegative while preserving the magnitude. So, sqrt(X^2)=|X|. This probably is the algebraic form you were looking for.